Probabilities of four statistically dependent discrete variable Let $A_0,A_1,B_0,B_1$ be random variables with a joint probability mass function, each of whom may get $1$ or $-1$. I have a few questions concerning them:


*

*Bell's inequality, here I have to show that:
$|E[A_0B_0]+E[A_0B_1]+E[A_1B_0]-E[A_1B_1]|\le 2$
Would this substitute as a sufficient proof:
$|E[A_0B_0]+E[A_0B_1]+E[A_1B_0]-E[A_1B_1]=|E[A_0(B_0+B_1)]+E[A_1(B_1-B_0)]|\le {(using \ triangle \ inequality)}\le |E[A_0(B_0+B_1)]|+|E[A_1(B_1-B_0)]|=???$
Now, if $P(B_0=1)=P(B_1=1)=1$, then $???=2$, else if $P(B_0=1)=P(B_1=-1)$, then also $???=2$. Where the probabilities of $A_0,A_1$ are uninteresting because $|E[A_i]|=1$.

*Next, one needs to show how $P(A_cB_d=(-1)^{cd})={1\over2}(1+(-1)^{cd}E[A_cB_d])$, where $c,d$ are indices which may take $0,1$. Again, as I understand, nothing can be inferred of the expectation value.

*One runs a trial with $A_c$ and $B_d$, assuming the probability of choosing each pair is $1/4$, what is the highest probability that in that trial run one gets $(1,-1)$ or $(-1,1)$ iff $c=d=1$, and $(1,1)$ or $(-1,-1)$ only when $c$ or $d$ equals $0$.

*Let $Z_{00},Z_{01},Z_{10},Z_{00}$ be four random variables which get $1$ or $-1$. Show that $E[Z_{00}Z_{10}]=E[Z_{01}Z_{11}]$ iff $Z_{ab}=X_aY_b$, where $X_A$ and $Y_b$ which may get $1$ or $-1$. I have only one direction:
$E[Z_{00}Z_{10}]=E[x_0y_0x_1y_0]=E[x_0x_1{y_0}^2]=E[x_0x_1]$
and
$E[Z_{01}Z_{11}]=E[x_0y_1x_1y_1]=E[x_0x_1{y_1}^2]=E[x_0x_1]$

*Assuming $E[Z_{00}Z_{10}]=E[Z_{01}Z_{11}]$, prove that when one discards the axiom that probability is a non-negative number then the inequality $|E[A_0B_0]+E[A_0B_1]+E[A_1B_0]-E[A_1B_1]|\le 2$ may be broken.


Thank you, whoever tries to help me! 
Any answer might help me, even if only for one of the items.
 A: Part 1. Let's calculate the expression $A_0B_0+A_0B_1+A_1B_0-A_1B_1$:
$$A_0B_0+A_0B_1+A_1B_0-A_1B_1=A_0(B_0+B_1)+A_1(B_0-B_1)$$
Now we have 2 cases: either $B_0=B_1$ or $B_0=-B_1$
In the 1st case we get $2A_0B_0$, in the 2nd case we get $2A_1B_0$. In both cases
$$A_0B_0+A_0B_1+A_1B_0-A_1B_1=\pm2$$
and it follows from here that
$$-2\leqslant E[A_0B_0+A_0B_1+A_1B_0-A_1B_1]\leqslant 2$$
A: Part 2.
$A_cB_d$ can assume values $1$ and $-1$.
Consider first the case that $c=1$ and $d=1$. Then,
$$P(A_1B_1 = 1) + P(A_1B_1 = -1) = 1$$
$$E[A_1B_1] = 1 \cdot P(A_1B_1 = 1) + (-1) \cdot  P(A_1B_1 = -1)$$
$$E[A_1B_1] = 1 - P(A_1B_1 = -1) - P(A_1B_1 = -1)$$
which implies,
$$P(A_1B_1 = -1) = \frac{1}{2}\bigg(1 - E[A_1B_1]\bigg)$$
Now consider (without loss of generality) $c=0$ and $d=1$.
$$P(A_0B_1 = 1) + P(A_0B_1 = -1) = 1$$
$$E[A_0B_1] = 1 \cdot P(A_0B_1 = 1) + (-1) \cdot  P(A_0B_1 = -1)$$
$$E[A_0B_1] = P(A_0B_1 = 1) - (1 - P(A_0B_1 = 1))$$
which implies,
$$P(A_0B_1 = 1) = \frac{1}{2}\bigg(1 + E[A_0B_1]\bigg)$$
The case is similar for $c=0, \ d=0$ and $c=1, \ d=0$
This has been generalized in 2 as $P\bigg(A_cB_d = (-1)^{cd}\bigg) = \frac{1}{2}\bigg(1 + (-1)^{cd}E[A_cB_d]\bigg)$
